A 2000 kg car is speeding at 72 km/h. Determine the retarding force of brakes required to stop it in 300m on a level road.
Answers
Answer:
20 n
Explanation:
Given,
Mass of car, M=2000kg
Velocity, V=72×
18
5
=20ms
−1
Apply kinematic equation of motion
v
2
−u
2
=2as
0−20
2
=2a×20
a=−10ms
−2
Breaking force, F=ma=2000×10=20 kN
Apply first kinematic equation
v=u+at
t=
a
v−u
=
−10
0−20
=2 sec
Breaking force is 20 kNand time is 2sec.
Answer:
Force (F) applied by brakes is 4000 N
Explanation:
Gɪᴠᴇɴ :-
Mass (m) = 2000 kg
Initial velocity (u) = 72 ×5/18 = 20m/s
Final velocity (v) = 36×5/18 = 10 m/s
ᴛᴏ ғɪɴᴅ :-
Force (F) applied by brakes
sᴏʟᴜᴛɪᴏɴ :-
On using 1st equation of motion, we get,
➦ v = u + at
→ 10 = 20 + a × 5
→ 10 - 20 = 5a
→ 5a = -10
→ a = -10/5
→ a = -2 m/s²
Hence,
Acceleration (a) = 2 m/s²
Negative sign shows deceleration or retardation
Now,
We know that,
➦ Force = Mass × Acceleration
➦ F = ma
➦ F = 2000 × -2
➦ F = -4000 N
Hence,
Force (F) applied by brakes is 4000 N
Negative sign shows that force must be applied in opposite direction of motion .